In the circuit shown in figure find the heat developed across each resistance in `2 s`.

Explanation

The `6Omega and 3Omega` resistances are in parallel. So, their combined resistance is
`1/R=1/6+1/3=1/2`
or `R=2Omega`
The equivalent simple circuit can be drawn as shown.

Current in the circuit `i=("netemf")/("total resistance")=20/(3+2+5)=2A`
`V=iR=(2)(2)=4V`
i.e. Potential difference across `6Omega and 3Omega` resistance are 4V. Now,
`H_(3Omega)` (which is connected in series) =`i^2Rt=(2)^2(3)(2)=24J`
`H_(6Omega)=V^2/Rt=(4)^(2)/(6) (2) = 16/3J`
`H_(3Omega)` (which is connected in parallel ) `=V^2/Rt=((4)^2(2))/3=32/3J`
and `H_(5Omega)=i^2Rt=(2)^2(5)(2)=40J`